A plane convex lens of refractive index $$1.5$$ and radius of curvature $$30$$ cm is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the object?

When one surface of a lens is silvered, it behaves as an equivalent mirror. The effective power of the system $$P_{\text{eq}}$$ is given by light passing through the lens, reflecting off the mirror, and passing through the lens again:

$$ P_{\text{eq}} = P_{\text{lens}} + P_{\text{mirror}} + P_{\text{lens}} = 2P_L + P_M $$

1. Power of the Lens ($$P_L$$)

Using the Lens Maker's Formula for a plano-convex lens ($$R_1 = \infty$$ for the plane surface, $$R_2 = -R$$ for the curved surface):

$$ P_L = \frac{1}{f_L} = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$

$$ P_L = (1.5 - 1)\left( \frac{1}{\infty} - \frac{1}{-30} \right) = (0.5) \left( \frac{1}{30} \right) = \frac{1}{60} \text{ cm}^{-1} $$

2. Power of the Mirror ($$P_M$$)

The silvered curved surface acts as a concave mirror.

$$ f_M = \frac{-R}{2} = \frac{-30}{2} = -15 \text{ cm} $$

$$ P_M = -\frac{1}{f_M} = -\left(\frac{1}{-15}\right) = \frac{1}{15} \text{ cm}^{-1} $$

3. Equivalent Power and Focal Length

$$ P_{\text{eq}} = 2\left(\frac{1}{60}\right) + \frac{1}{15} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10} \text{ cm}^{-1} $$

The equivalent focal length $$F_{\text{eq}}$$ is:

$$ F_{\text{eq}} = -\frac{1}{P_{\text{eq}}} = -10 \text{ cm} $$

The system behaves like a concave mirror of focal length $$10 \text{ cm}$$.

4. Finding Object Distance

To obtain a real image of the same size as the object (magnification $$m = -1$$), the object must be placed at the center of curvature of the equivalent mirror.

$$ u = 2F_{\text{eq}} = 2(10) = 20 \text{ cm} $$

Answer: Option (A) $$20 \text{ cm}$$